Chemistry: Post your doubts here!

[iKhaled]

bro .. i posted the WRONG year !
its 2004 .. sorry for that ... http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s04_qp_1.pdf
q8 .... i kept thinking from wr the calcium any way can u help me out with this

haha anyway its okay..

question (8)

you will need to make a hess cycle..see the enthalpy change of formation is when one mole of a compound is formed from its elements int heir standart conditions under standard conditions so we have formed one mole of aluminium oxide and 3 moles of iron (iii) oxide so here is the cycle

question 28

see u have been given 2 hints here. 1 is that that the compound is unreactive to oxidising agents and 2 is that it has been dehydrated so obviously its a third alcohol..now look at the compound which had a hydroxyl group attached to a carbon which is attached to 2 other carbons and the only compound which has this is D! because if u remove the double bond and add the H and OH instead u will see that the OH is tertiary..did u get it ?

 

[iKhaled]

yep bro!
thanks

anytime

 

[ahmed abdulla]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s05_qp_1.pdf
q 2 ..

 

[AbbbbY]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s05_qp_1.pdf
q 2 ..

B!

2 mols of NaN3 give 2 mols of Na and 3 mols of N2

So, 1 mol of NaN3 will give 1 mol of Na and 1.5 mol of N2

10 mol of Na gives 1 mol of N2
So 1 mol of Na will give 0.1 mol of N2

Hence total N2 given is 1.6mol

 

[AbbbbY]

Is the unit for 1000/reaction time in seconds , s^-1 or 1000s^-1 ?

s^-1

 

[Rahma Abdelrahman]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s11_qp_12.pdf
Please help with the following questions:
Q12
Q26, why 6 isomers, why not 5?
Q29 (If u can explain it here , if not no problemo )
Q30, Why answer is A and not B?
Q35.... In the first sentence, how does N undergo REDOX rx. ?? I guess it is only oxidised as oxidation state increases from -3 to zero
Thanks in advance..

Please someone reply! This is the third time I post these doubts

 

[ahmed abdulla]

B!

2 mols of NaN3 give 2 mols of Na and 3 mols of N2

So, 1 mol of NaN3 will give 1 mol of Na and 1.5 mol of N2

10 mol of Na gives 1 mol of N2
So 1 mol of Na will give 0.1 mol of N2

Hence total N2 given is 1.6mol

but didnt they say for NaN3 .. ? y we used the total moles produced of N2 ?

 

[AbbbbY]

but didnt they say for NaN3 .. ? y we used the total moles produced of N2 ?

It's a chain reaction so you have to consider both products since they're happening simultaneously. Getit?

 

[ahmed abdulla]

It's a chain reaction so you have to consider both products since they're happening simultaneously. Getit?

yep

 

[ahmed abdulla]

can someone help me with HESS LAW ... i always get stuck here in drawing , in getting answers ? ...

Q ) Iodine trichloride, ICl3, is made by reacting iodine with chlorine.
I2(s) + Cl2(g) → 2ICl(s) ; ΔH o = +14 kJ mol–1
ICl(s) + Cl2(g) → ICl3(s) ; ΔH o = –88 kJ mol–1
By using the data above, what is the enthalpy change of the formation for solid iodine trichloride?
A –60 kJ mol–1
B –74 kJ mol–1
C –81 kJ mol–1
D –162 kJ mol–1

AbbbbY

 

[AbbbbY]

can someone help me with HESS LAW ... i always get stuck here in drawing , in getting answers ? ...

Q ) Iodine trichloride, ICl3, is made by reacting iodine with chlorine.
I2(s) + Cl2(g) → 2ICl(s) ; ΔH o = +14 kJ mol–1
ICl(s) + Cl2(g) → ICl3(s) ; ΔH o = –88 kJ mol–1
By using the data above, what is the enthalpy change of the formation for solid iodine trichloride?
A –60 kJ mol–1
B –74 kJ mol–1
C –81 kJ mol–1
D –162 kJ mol–1

AbbbbY

ΔHf = ΔHp - ΔHr


You don't need the Hess Law for this.

Since enthalpy of formation is formation of 1 mol from constituent elements (which is a two-step reaction in this case)

So, ΔHf for 2ICl = +14, so ICl = +7

Add that to the second step you have your enthalpy change (In that, -88+7 = -81 thus C)

Wait a few mins I'll upload a handout on Hess's Law. It's very very easy.

 

[AbbbbY]

can someone help me with HESS LAW ... i always get stuck here in drawing , in getting answers ? ...

Q ) Iodine trichloride, ICl3, is made by reacting iodine with chlorine.
I2(s) + Cl2(g) → 2ICl(s) ; ΔH o = +14 kJ mol–1
ICl(s) + Cl2(g) → ICl3(s) ; ΔH o = –88 kJ mol–1
By using the data above, what is the enthalpy change of the formation for solid iodine trichloride?
A –60 kJ mol–1
B –74 kJ mol–1
C –81 kJ mol–1
D –162 kJ mol–1

AbbbbY

These three should cover AS and A2 energetics sufficiently. If you don't get it, let me know what you don't get and I'll explain.

 

[omarjaved619]

HELLLPPPPP!

S12_qp11 Question 1
Why is the answer BF3 and NOT CH3-? :S

And same paper, question 14. Thanks a ton!

 

[darknessinme]

Please someone reply! This is the third time I post these doubts

12. Going down periodic table, electronegativity decreases. Going right, electronegativity increases. So go right from Be, it increases, and go down one it decreases. These cancel so it's electronegativity is similar to Al.
Answer B

13. Did you consider the cis and trans isomers?

29. Two -OH group's add at the double bond with cold,dilute KMnO4. The carbon atoms attached to the -OH groups both become chiral, so +2.
With hot, conc. KMnO4, double breaks, and ketone group and carboxylic acid group form. The carbons still not chiral. The -OH group on the very left of the original molecule is oxidised to ketone, so that carbon becomes non-chiral, so -1.
Answer D.

30. B is acid hydrolysis of a nitrile. You would get ethanoic acid. Question says which would not give. A is base hydrolysis of nitrile, where you get the sodium salt sodium ethanoate.
Answer A.

 

[ahmed abdulla]

ΔHf = ΔHp - ΔHr


You don't need the Hess Law for this.

Since enthalpy of formation is formation of 1 mol from constituent elements (which is a two-step reaction in this case)

So, ΔHf for 2ICl = +14, so ICl = +7

Add that to the second step you have your enthalpy change (In that, -88+7 = -81 thus C)

Wait a few mins I'll upload a handout on Hess's Law. It's very very easy.

thanks aloot bro ... now at the end of the AS year .. i dont think i need this for paper 1 ! it takes time

 

[ahmed abdulla]

ΔHf = ΔHp - ΔHr


You don't need the Hess Law for this.

Since enthalpy of formation is formation of 1 mol from constituent elements (which is a two-step reaction in this case)

So, ΔHf for 2ICl = +14, so ICl = +7

Add that to the second step you have your enthalpy change (In that, -88+7 = -81 thus C)

Wait a few mins I'll upload a handout on Hess's Law. It's very very easy.

as per your equation .. p-r .... shouldnt it be -88-(+7) =-95 ... or did u add them ?

 

[AbbbbY]

thanks aloot bro ... now at the end of the AS year .. i dont think i need this for paper 1 ! it takes time

For P1, simple use ΔHf = ΔHp - ΔHr and ΔHc = ΔHr - ΔHp (I remember it from Crap. Crap = Cr-p. And thus, F is the other way round).

 

[AbbbbY]

as per your equation .. p-r .... shouldnt it be -88-(+7) =-95 ... or did u add them ?

Enthalpy of formation is given for both the reactions. If the enthalpy of formation of individual products was given we'd have done that.
Basically, they've already done the p-r and have given the value (+14 and -88 respectively)

 

[AbbbbY]

HELLLPPPPP!

S12_qp11 Question 1
Why is the answer BF3 and NOT CH3-? :S

And same paper, question 14. Thanks a ton!

1- BF3 =
Each F has 7 outer electrons
Each B has 3 outer electrons
So, B forms 3 bonds and hence has 6 electrons (incomplete octet!)

CH3-
Each C has 4 outer electrons
Each H has one outer electron
CH3 has an overall negative charge.

So, C forms 3 bonds and will have 6 electrons.
One electron is left but there is a negative charge so it is gaining an electron. [Well that's what my teacher told me back when I asked him!]


14- 4Al + Ba(NO3)2 -> 2(Al2O3) + Ba + N2

1 mol : 1 mol
0.783/(137+62(2)) = 0.003 mol

1 mol occupies 24dm3
0.003 mol will occupy 0.072dm3
Hence,B

 

[Rahma Abdelrahman]

12. Going down periodic table, electronegativity decreases. Going right, electronegativity increases. So go right from Be, it increases, and go down one it decreases. These cancel so it's electronegativity is similar to Al.
Answer B

13. Did you consider the cis and trans isomers?

29. Two -OH group's add at the double bond with cold,dilute KMnO4. The carbon atoms attached to the -OH groups both become chiral, so +2.
With hot, conc. KMnO4, double breaks, and ketone group and carboxylic acid group form. The carbons still not chiral. The -OH group on the very left of the original molecule is oxidised to ketone, so that carbon becomes non-chiral, so -1.
Answer D.

30. B is acid hydrolysis of a nitrile. You would get ethanoic acid. Question says which would not give. A is base hydrolysis of nitrile, where you get the sodium salt sodium ethanoate.
Answer A.

12--> I got ur point, but I assumed its Mg as its closest to Al in the periodic table..
(Its 26, not 13) --> Oh.. I guess not thanks for reminding me
29--> Got it..
30--> Yeah..
Thanks ..
Sorry but what about Q35 ?